First slide

Questions on calorimetry without phase change

Question

A liquid of mass M and specific heat S is at a temperature 2 t. If another liquid of mass M and thermal capacity 1.5 times, at a temperature of t/3 is added to it, the resultant temperature will be:

Moderate

A

4t/3
B

t
C

t/2
D

2t/3

Solution

Let the common temperature of the mixture be T.
For first liquid, mass, M, specific heat of liquid, S and initial temperature of liquid, 2t.
For second liquid, mass, M, specific heat of liquid, 1.5 S and initial temperature of liquid, $\frac{t}{3}$
Heat loss by the hot liquid = Heat consumed by cold liquid
$MS (2 t - T) = M (1 .5 S) (T - \frac{t}{3})$
$\Rightarrow 2 t - T = 1 .5 T - \frac{t}{2}$
$\Rightarrow \frac{5}{2} T = \frac{5}{2} t$
$\Rightarrow T = t$

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